%I #18 Sep 05 2018 06:15:46
%S 1,8,64,1,512,16,4096,192,1,32768,2048,24,262144,20480,384,1,2097152,
%T 196608,5120,32,16777216,1835008,61440,640,1,134217728,16777216,
%U 688128,10240,40,1073741824,150994944,7340032,143360,960,1,8589934592,1342177280,75497472,1835008,17920,48
%N Triangle read by rows: T(0,0) = 1; T(n,k) = 8 * T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.
%C The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013615 ((1+8*x)^n) and along skew diagonals pointing top-right in center-justified triangle given in A038279 ((8+x)^n).
%C The coefficients in the expansion of 1/(1-8x-x^2) are given by the sequence generated by the row sums.
%C The row sums are Denominators of continued fraction convergents to sqrt(17), see A041025.
%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 8.12310562561766054982... (a metallic mean), when n approaches infinity (see A176458: (4+sqrt(17))).
%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, Pages 70, 98
%H Zagros Lalo, <a href="/A317028/a317028.pdf">Left-justified triangle</a>
%H Zagros Lalo, <a href="/A317028/a317028_1.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (1 + 8x)^n</a>
%H Zagros Lalo, <a href="/A317028/a317028_2.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (8 + x)^n</a>
%e Triangle begins:
%e 1;
%e 8;
%e 64, 1;
%e 512, 16;
%e 4096, 192, 1;
%e 32768, 2048, 24;
%e 262144, 20480, 384, 1;
%e 2097152, 196608, 5120, 32;
%e 16777216, 1835008, 61440, 640, 1;
%e 134217728, 16777216, 688128, 10240, 40;
%e 1073741824, 150994944, 7340032, 143360, 960, 1;
%e 8589934592, 1342177280, 75497472, 1835008, 17920, 48;
%e 68719476736, 11811160064, 754974720, 22020096, 286720, 1344, 1;
%e 549755813888, 103079215104, 7381975040, 251658240, 4128768, 28672, 56;
%e 4398046511104, 893353197568, 70866960384, 2768240640, 55050240, 516096, 1792, 1;
%t t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 8 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 11}, {k, 0, Floor[n/2]}] // Flatten
%o (PARI) T(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, 8*T(n-1, k)+T(n-2, k-1)));
%o tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n, k), ", ")); print); \\ _Michel Marcus_, Jul 20 2018
%Y Row sums give A041025.
%Y Cf. A013615, A038279, A176458.
%Y Cf. A001018 (column 0), A053539 (column 1), A081138 (column 2), A140802 (column 3), A172510 (column 4).
%K tabf,nonn,easy
%O 0,2
%A _Zagros Lalo_, Jul 19 2018